There $x_1,x_2,x_3,x_4\in\mathbb R$ s.t. each one of them subtracted from the arithmetic mean of the rest of them gives the result(respectively):$-8,4,8,-4$. How many solutions are there?
My attempt:
$$\begin{cases}\frac{x_1+x_2+x_3}{3}-x_4=-8\\\frac{x_1+x_2+x_4}{3}-x_3=4\\\frac{x_1+x_3+x_4}{3}-x_2=8\\\frac{x_2+x_3+x_4}{3}-x_1=-4\end{cases}$$
I saw we can add $3$ arbitrary equations to the $4^{\text{th}}$ one and annulate one row/equation. Now there are $3$ equations left and $4$ unknowns$\implies$ there are infinitely many solutions.
$$\begin{bmatrix}1&1&1&-3&|&-24\\1&1&-3&1&|&12\\1&-3&1&1&|&24\\-3&1&1&1&|&-12\end{bmatrix}\sim\begin{bmatrix}1&1&1&-3&|&-24\\1&1&-3&1&|&12\\1&-3&1&1&|&24\\0&0&0&0&|&0\end{bmatrix}$$ $$\rightarrow\begin{bmatrix}1&1&1&-3&|&-24\\1&1&-3&1&|&12\\1&-3&1&1&|&24\end{bmatrix}\sim\begin{bmatrix}1&1&1&-3&|&-24\\0&0&-4&4&|&36\\0&-4&0&4&|&48\end{bmatrix}\sim\begin{bmatrix}1&1&1&-3&|&-24\\0&0&-1&1&|&9\\0&-1&0&1&|&12\end{bmatrix}\sim\begin{bmatrix}1&1&0&-2&|&-15\\0&0&-1&1&|&9\\0&-1&0&1&|&12\end{bmatrix}\sim\begin{bmatrix}1&0&0&-1&|&-3\\0&0&-1&1&|&9\\0&-1&0&1&|&12\end{bmatrix}$$ $x_4$ appears in every equation, so I expressed the remaining unknowns over it: $$\begin{cases}x_1-x_4=-3\implies x_1=x_4-3\\x_3-x_4=-9\implies x_3=x_4-9\\x_2-x_4=-12\implies x_2=x_4-12\end{cases}$$ My solution: $$[x_4-3,x_4-12,x_4-9,x_4]^{\tau}$$
Is this correct?
Let $s:=\dfrac{x_1+x_2+x_3+x_4}3$, and
$$\begin{cases}\dfrac{x_1+x_2+x_3}{3}-x_4=-8\\\dfrac{x_1+x_2+x_4}{3}-x_3=4\\\dfrac{x_1+x_3+x_4}{3}-x_2=8\\\dfrac{x_2+x_3+x_4}{3}-x_1=-4\end{cases}$$
can be rewritten$$\begin{cases}s-\dfrac{4x_4}3=-8\\s-\dfrac{4x_3}3=4\\s-\dfrac{4x_2}3=8\\s-\dfrac{4x_1}3=-4.\end{cases}$$
By summing the equations we obviously have $4s-4s=0$, and we can solve for the four unknowns in terms of $s$.